Deriving the Law of CosinesDate: 04/01/98 at 01:04:26 From: Brooke Lawrenson Subject: Geometry I wonder if the Pythagorean Theorem will work with a non-right triangle and how to go about proving it. Date: 04/01/98 at 11:47:56 From: Doctor Rob Subject: Re: Geometry This is a good question, and it shows good thinking. The Pythagorean Theorem will not work for triangles that are not right triangles. When you have a triangle which is not right, the closest analogue is something called the Law of Cosines. If the sides of a triangle are labeled a, b, and c, and the angles opposite them A, B, and C, respectively, then: c^2 = a^2 + b^2 - 2*a*b*cos(C) Here "cos(C)" means the cosine of the angle C. If the triangle is a right triangle with right angle C, then the cosine of C, that is, the cosine of 90 degrees, is zero, so you get c^2 = a^2 + b^2, the Pythagorean Theorem. If the angle is other than 90 degrees; however, the cosine of C is not zero. For example, if the angle C is 60 degrees, its cosine is 1/2, and you get the equation c^2 = a^2 +b^2 - a*b. If C is 120 degrees, its cosine is -1/2, and you get the equation c^2 = a^2 + b^2 + a*b. You will learn about cosines and prove the Law of Cosines when you study trigonometry. The proof depends on the Pythagorean Theorem, strangely enough! Draw triangle ABC with sides a, b, and c, as above. Drop a perpendicular from A to BC, meeting it at point P. Let the length AP be x, and the length CP be y. Then BP = a-y. Then apply the Pythagorean Theorem to the two triangles ACP and ABP: CP^2 + AP^2 = AC^2 or y^2 + x^2 = b^2 BP^2 + AP^2 = AB^2 or (a-y)^2 + x^2 = c^2 Now subtract the first equation from the second, to eliminate x: (a-y)^2 - y^2 = c^2 - b^2 a^2 - 2*a*y + y^2 - y^2 = c^2 - b^2 a^2 - 2*a*y = c^2 - b^2 c^2 = a^2 + b^2 - 2*a*b*(y/b) Now y/b is just the cosine of angle C (if you don't know this, trust me), so: c^2 = a^2 + b^2 - 2*a*b*cos(C) Observe that C is a right angle if and only if y = 0 and points P and C coincide. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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